1. ## Prove set theorems

For any sets A, B, C in a universe U:

Prove that:

If A n B = C n B and A n B' = C n B' then A = C

the "n" symbol means "intersect"

2. Hello,
Originally Posted by mbcsantin
For any sets A, B, C in a universe U:

Prove that:

If A n B = C n B and A n B' = C n B' then A = C

the "n" symbol means "intersect"
$(A \cap B) \cup (A \cap B')=(C \cap B) \cup (C \cap B')$
$A \cap (B \cup B')=C \cap (B \cup B')$, by associative law (Algebra of sets - Wikipedia, the free encyclopedia)

But $B \cup B'=U$ by definition of the complement. And every set intersected with the universe will result in the set itself.

Hence $A=C$

3. Originally Posted by Moo
Hello,

$(A \cap B) \cup (A \cap B')=(C \cap B) \cup (C \cap B')$
$A \cap (B \cup B')=C \cap (B \cup B')$, by associative law (Algebra of sets - Wikipedia, the free encyclopedia)

But $B \cup B'=U$ by definition of the complement. And every set intersected with the universe will result in the set itself.

Hence $A=C$
Thanks!
But just a question..

So this is what I'm trying to prove:
If A n B = C n B and A n B' = C n B' then A=C

(AnB)U(AnB')=(CnB)U(CnB')

My question is, wouldn't it be
(A n B) U (C n B) = (A n B') U (C n B') instead??

4. This it true for $\left( {\forall X,Y} \right)\left[ {X = \left( {X \cap Y} \right) \cup \left( {X \cap Y'} \right)} \right]$.

So
$\begin{array}{rcl}
A & = & {\left( {A \cap B} \right) \cup \left( {A \cap B'} \right)} \\
{} & = & {\left( {C \cap B} \right) \cup \left( {C \cap B'} \right)} \\
{} & = & C \\
\end{array}$

5. Originally Posted by mbcsantin
Thanks!
But just a question..

So this is what I'm trying to prove:
If A n B = C n B and A n B' = C n B' then A=C